document.write( "Question 616937: car A travels x km for every litre of petrol.while car B travels (x+5)km for every litre of petrol. if car A uses 4 litres of petrol more than car B in covering 400km, find the number of litres of petril used by car B for the journey. \n" ); document.write( "
Algebra.Com's Answer #388184 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! car A travels x km for every litre of petrol.while car B travels (x+5)km for every litre of petrol. \n" ); document.write( "if car A uses 4 litres of petrol more than car B in covering 400km, find the number of litres of petril used by car B for the journey. \n" ); document.write( ": \n" ); document.write( "Car A's consumption - Car B consumption = 4 liters \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "multiply by x(x+5) to clear the denominators, results: \n" ); document.write( "400(x+5) - 400x = 4x(x+5) \n" ); document.write( "400x + 2000 - 400x = 4x^2 + 20x \n" ); document.write( ": \n" ); document.write( "Combine like terms, A quadratic equation \n" ); document.write( "4x^2 + 20x - 2000 = 0 \n" ); document.write( "; \n" ); document.write( "simplify, divide by 4 \n" ); document.write( "x^2 + 5x - 500 = 0 \n" ); document.write( ": \n" ); document.write( "Factors to \n" ); document.write( "(x+25)(x-20) = 0 \n" ); document.write( ": \n" ); document.write( "the positive solution \n" ); document.write( "x = 20 km/liter for Car A \n" ); document.write( "then \n" ); document.write( "25 km/liter for Car B \n" ); document.write( "therefore \n" ); document.write( "Car B: \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Confirm this solution by finding the consumption of Car A \n" ); document.write( "400/20 = 20 liters, 4 liters more \n" ); document.write( " |